06-07-2025

9610WJA801936N250002 JA

PHYSICS

SECTION-I

1) If then angle between will be

(A) 90°
(B) 120°
(C) 0°
(D) 60°

2) Sum of two unit vectors is a unit vector. What is magnitude of their vector difference?

(A)
(B)

(C)

(D)

3) A vector is directed along 30° west of north direction and another vector along 15° south of
east. Their resultant cannot be in ________ direction.

(A) North
(B) East
(C) North-East
(D) South

4) Calculate the area of the parallelogram when adjacent sides are given by the vectors
and

(A)
(B)
(C)
(D)

5) Find the value of m so that the vector may be perpendicular to the vector

(A) 2
(B) 4
(C) 6
(D) 8

6) The vector are and The angle between the two vectors will be:

(A)

(B)

(C)

(D)

7) Projection of vector on vector is

(A)
(B)

(C)

(D)

8) When the speed of a car is u, the minimum distance over which it can be stopped is s. If the speed
becomes nu, what will be the minimum distance over which it can be stopped?

(A) s/n
(B) ns
(C) s/n2
(D) n2s

9) A thief is running away on a straight road in a jeep moving with a speed of A policeman
chases him on a motor cycle moving at a speed of If the instantaneous separation of the
jeep from the motor cycle is 100 m, how long will it take for the policeman to catch the thief?

(A) 1 s
(B) 19 s
(C) 90 s
(D) 100 s

10) The acceleration of a particle which moves along the positive x-axis varies with its position as
shown in figure. If the velocity of the particle is at x = 0, then velocity of the particle at x =
1.4m is (in ms-1)
(A) 1.6
(B) 1.2
(C) 1.4
(D) None of these

11) A particle is moving along the x-axis whose acceleration is given by where x is the
location of the particle. At t = 0, the particle is at rest at The distance travelled by the
particle in 5s is

(A) Zero
(B) 42 m
(C) Infinite
(D) None of these.

12) The velocity-time graph for a particle moving along a straight line is given in each situation of
column I. Match the graph in column-I with corresponding statements in column-II.

Column-I Column-II

i. a. Speed of particle is continuously decreasing

Magnitude of acceleration of particle is

ii. b.
decreasing with time.

Direction of acceleration of particle does not

iii. c.
change

Magnitude of acceleration of particle does

iv. d.
not change
 particle will never come back to its initial
e.
position.
(A)
(B)
(C)
(D)

13) The velocity-time graph of a car moving on a straight track is given below. The car weighs 1000
kg.

The distance travelled by the car during the whole motion is:

(A) 50 m
(B) 75 m
(C) 100 m
(D) 150 m

14) The displacement-time graph of a moving particle with constant acceleration is shown in the
figure. The velocity-time graph is best given by:

(A)

(B)
(C)

(D)

15) The graph to the right is a plot of position versus time. For which labelled region is the velocity
positive and the acceleration negative?

(A) a
(B) b
(C) c
(D) d

16) The graph shows position as a function of time for two trains running on parallel tracks. Which
one of the following statement true?

(A) At time tB, both trains have the same velocity.

(B) Both trains have the same velocity at some time after tB
(C) Both trains have the same velocity at some time before tB
(D) Somewhere on the graph, both trains have the same acceleration.
17) Each of the three graphs represents acceleration versus time for an object that already has a
positive velocity at time t1. Which graphs show an object whose speed is increasing for the entire
time interval between t1 and t2?

(A) graph I, only

(B) graph I and II, only
(C) graph I and III, only
(D) graphs I, II and III

18) Graph of (1/v) vs x for a particle under motion is as shown, where v is velocity and x is position.
The time taken by particle to move from x = 4m to x = 12m is:

(A)

(B) 10 sec
(C) 8 sec
(D) 12 sec

19) The velocity (v) and time (t) graph of a body in a straight line motion is shown in the figure. The
point S is at 4.333 seconds. The total distance covered by the body in 6s is:

(A) 12m

(B)

(C) 11m

(D)

20) A stone is thrown vertically upward with an initial velocity u from the top of a tower, reaches the
ground with a velocity 3u. The height of the tower is:-
(A)

(B)

(C)

(D)

SECTION-II

1) A ball is thrown vertically upwards. Its distance s from a fixed point varies with time t according
to the following graph. Calculate velocity of projection of the ball. (Take g = 10 m/s2)

2) If the angle Between will be P is-

3) A body starts from rest and moves for 'n' seconds with uniform acceleration 'a'. its velocity after n
seconds is 8 m/s. The displacement of the body in last 3 seconds is 15 m. Find a (in m/s2).

4) A car is moving with 54 km/h speed at t = 0. Its brakes produce retardation of 2.5 m/s2. If the
driver applies brakes till the car stops then the displacement of car is (in m).

5) A particle moves such that its position is given by where x is in metre and t is in
second. Find its average velocity (in m/s) from t = 2 s to t = 5 s.

CHEMISTRY

SECTION-I

1) Bohr's atomic model can explain the spectrum of

(A) Hydrogen atoms only

(B) Atoms or ions which are unielectron
(C) Atom or ions which have only two electrons
(D) Hydrogen molecule

2) The wave number of the first line of Balmer series of hydrogen is 15200 cm-1. The wave number of
the first Balmer line of Li2+ ion is

(A) 15200 cm-1

(B) 60800 cm-1
(C) 76000 cm-1
(D) 136800 cm-1

3) The radius of an atomic nucleus is of the order of

(A)
(B)
(C)
(D)

4) When electromagnetic radiation of wavelength 310 nm fall on the surface of sodium, electrons are
emitted with maximum K.E. = 1.5 eV. Determine the work function (W0) of sodium.

(A) 2.5 eV
(B) 5 eV
(C) 10 eV
(D) 12.5 eV

5) In the first Bohr orbit of H atom the energy of an electron is -13.6 eV. The possible energy
value of excited state for electron in an orbit of hydrogen atom is :

(A) -3.4 eV
(B) -4.2 eV
(C) -8.6 eV
(D) +6.8 eV

6) According to Bohr's atomic theory :-

(1) Kinetic energy of electron is .
2
(2) The velocity (v) of electron is Z.

(3) Frequency of revolution of electron in an orbit is .

(4) Coulombic force of attraction on the electron is .

Choose the most appropriate answer from the options given below :

(A) (3) only

(B) (1) only
(C) (1), (3) and (4) only
(D) (1) and (4) only

7) The shortest wavelength of H atom in the Lyman series is . The longest wavelength in the
Balmer series of He+ is :-

(A)

(B)

(C)

(D)

8) An α-particle accelerated through V volt is fired towards a nucleus. It distance of closest approach
is r. If a proton accelerated through the same potential is fired towards the same nucleus, the
distance of closest approach of the proton will be :

(A) r
(B) 2r

(C)

(D)

9) Assume that 10-17 J of light energy is needed by the interior of the human eye to see an object.
How many photons of green light ( λ=495nm) are needed to generate this minimum energy

(A) 25
(B) 30
(C) 35
(D) 40

10) For which of the following species, Bohr model is not valid:

(A) He+
(B) H
(C) Li2+
(D) B3+

11) The ratio of radius of two different orbits in a H-atom is 4 : 9, then the ratio of the frequency of
revolution of electron in these orbits is:

(A) 2:3
(B) 27:8
(C) 3:2
(D) 8:27

12) When an electron is brought close to the nucleus of the atom from an infinite distance, the
energy of the electron-nucleus system changes such that it

(A) Increases to a smaller negative value

(B) Increases to a greater positive value
(C) Decreases to a smaller positive value
(D) Decreases to a greater negative value

13) Graph of incident frequency with stopping potential (Vs) in photoelectric effect is

(A)

(B)

(C)

(D)

14) The third line in Balmer series corresponds to an electronic transition between which Bohr's
orbits in hydrogen:

(A) 5 3
(B) 5 2
(C) 4 3
(D) 4 2
15) The ratio of difference in wavelengths of 1st and 2nd lines of Lyman series in H-like atom to
difference in wavelength for 2nd and 3rd lines of same series in same species is:

(A) 2.5:1
(B) 3.5:1
(C) 4.5:1
(D) 5.5:1

16) The simplest radius ratio of 3rd orbit of Li+2 and fourth orbit of C5+ ion is , Then the value of
(x+y) is

(A) 15
(B) 16
(C) 17
(D) 18

17) The kinetic and potential energy (in eV) of electron present in third Bohr's orbit of hydrogen
atom are respectively

(A) –1.51, –3.02

(B) 1.51, –3.02
(C) 3.02, –1.51
(D) 1.51, –1.51

18) Light of wavelength λ shines on a metal surface with intensity X, and the metal emits Y electrons
per second of average energy Z. What will be changes observed in Y and Z if X is double?

(A) Y will be double and Z will became half

(B) Y will remain same and Z will be doubled
(C) Both Y and Z will be double
(D) Y will be double but Z will remain same.

19) If the velocity of 2nd Bohr orbit of H is 'a'. Then the velocity of first Bohr orbit of He+ is.

(A) 2a

(B)

(C) 4a

(D)

20) The mass to charge ratio for A+ ions is 1.97×10-7 kg C–1. Calculate the mass of A atom.

(A) 3.155×10–26 kg
(B) 3.155×1029 kg
(C) 31.55×1026kg
(D) 6.155×1028kg

SECTION-II

1) A certain transition in H spectrum from an excited state to ground state in one or more steps
gives rise to total 10 lines. How many of these belong to UV spectrum ?

2)

A certain dye absorbs lights of = 400 nm and then fluorescence light of wavelength 500 nm.
Assuming that under given condition 40% of the absorbed energy is re-emitted as fluorescence,
calculate the ratio of quanta absorbed to number of quanta emitted out.

3) Calculate the value of A.

where Energy of electron in nth orbit, Z = atomic number of hydrogen / hydrogen

like species.

4) Photon having wavelength 310 nm is used to break the bond of A2 molecule having bond energy
288 kJ mol–1 then % of energy of photon converted to the K.E is [ hc = 12400 eV Å, 1 eV/atom = 96
kJ/mol]

5) The energy of an electron in an excited Li+2 ion is –13.6 eV. If angular momentum of electron in
that state is , then the value of 'X' is (h = 6.625 × 10–34 J. sec)

MATHEMATICS

SECTION-I

1) The nth term of the series is 9901. The value of n is

(A) 100
(B) 90
(C) 900
(D) 99

2) upto is

(A) 4
(B) 5
(C) 6
(D) 7
3) Let G be the geometric mean of two positive numbers a and b, and M be the arithmatic mean

of and If : G is 4 : 5, then a : b can be :

(A) 2 : 3
(B) 1 : 4
(C) 1 : 2
(D) 3 : 4

4) The sum of the first 20 terms common between the series 3 + 7 + 11 + 15 + ......... and 1 + 6 +
11 + 16 + .......... is :

(A) 4220
(B) 4020
(C) 4000
(D) 4200

5) Consider a decreasing G.P. : g1, g2, g3, ...... gn ....... such that g1 + g2 + g3 = 13 and
then which of the following does not hold?

(A) The greatest term of the G.P. is 9

(B) 3g4 = g3
(C) g1 = 1
(D) g2 = 3

6) If is the G.M. between distinct positive numbers a and b, then the value of n is

(A) 0
(B) 1
(C) 1/2
(D) None of these

7) If a, b, c are in H.P., then are in

(A) A.P.
(B) G.P.
(C) H.P.
(D) A.G.P

8) For a sequence and . Then is

(A)
(B)

(C)

(D)

9) If 16x2 + 4y2 + z2 = 2xyz ; then x, y, z are in :-

(A) A.P.
(B) G.P.
(C) H.P.
(D) None of these

10) If H1, H2, ...., H20 are 20 harmonic means between 2 and 3, then

(A) 20
(B) 21
(C) 40
(D) 38

11) Let an be the nth term of a G.P. of positive numbers. Let and , such
that , then the common ratio is

(A)
(B)
(C)
(D)

12) A sequence is defined by x1 = 2 and for all . Find x101

(A)

(B)

(C)

(D)

13) The sum is equal to

(A)

(B)

(C)

(D)

14) Four numbers are in arithmetic progression. The sum of first and last terms is 8 and the product
of both middle terms is 15. The least number of the series is.

(A) 4
(B) 3
(C) 2
(D) 1

15) If then n = ___

(A) 14
(B) 7
(C) 15
(D) 17

16) The interior angles of a polygon are in A.P. If the smallest angle is 100° and the common
difference is 4°, then the number sides is

(A) 5
(B) 7
(C) 36
(D) 44

17) Find .

(A) 1023
(B) 6138
(C) 2046
(D) 3069

18) The greatest value of p = a b c , when a + b + c = 18, is (a, b, c ∈ R+) -

2 3 4

(A) 44.64.84
(B) 42.63.84
(C) 32.63.84
(D) 42.64.82

19) is

(A)

(B)

(C)

(D)

20) For any x, y R, xy > 0 then the minimum value of equals

(A)
(B) 2
(C)
(D) 3

SECTION-II

1) Sum of twenty terms of the series 1 + (1 + 2) + (1 + 2 + 3) + .... is

2) If Sn denotes the sum of the first n terms of an AP and if then

3) There are n AM 's between 3 and 54 such that 8th mean : (n–2)th mean = 3 : 5 then value of n is

4) Let S denote the set of all real values of x such that

, then the number of elements in S is

5) Minimum value of (for real positive numbers a, b, c) is

 ANSWER KEYS

PHYSICS

SECTION-I

Q. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
A. C B D B C D B D D B A B B A D C D B D B

SECTION-II

Q. 21 22 23 24 25
A. 20 2 2 45 8

CHEMISTRY

SECTION-I

Q. 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
A. B D B A A C C A A D B D A B B C B D C A

SECTION-II

Q. 46 47 48 49 50
A. 4 2 8 25 34

MATHEMATICS

SECTION-I

Q. 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70
A. A C B B C C C B B C A C B D B A B B B B

SECTION-II

Q. 71 72 73 74 75
A. 1540 6 16 1 6
 SOLUTIONS

PHYSICS

1)

Resultant of two vectors can be given by

2)

3)
The direction of resultant must lie in between smaller angle between therefore
resultant in south direction is not possible.

4) We know that area of the parallelogram is equal to magnitude of the cross product of given
vectors. Now,

5) The given vectors will be perpendicular if their dot product is zero.

or
or
6)

7)

8)

When the initial velocity is made n times, the distance over which it can be stopped becomes
n2 times.

9)

Relative velocity of policeman w.r.t. the thief is

Since the relative separation between them is 100 m, the time taken will be = relative
separation/relative velocity = 100/1 = 100 s.

10)

11)

As both the particle velocity and acceleration are zero at t = 0, it will always remain at rest
and hence distance travelled at any time interval would be zero.

12)

In case A and B, acceleration is constant but speed first decreases and then increases.
In case C and D, the velocity does not change sign. Hence, direction of acceleration is
constant. Speed and magnitude of acceleration decrease with time.

13)

Distance travelled by the car in 10 second is equal to displacement in 10 second and it is same
as area under the v-t curve.
14)

From figure: (given in problem)

For time interval t = 0 to t = 1 sec
Slope of x-t graph is negative and increasing, so velocity increases in negative direction.
For t = 1 to 2 sec
The slope is +ve and decreasing, so velocity is decreasing in +ve direction and become zero at
t=2
So, (1) is correct.

15)

During 'd' slope is +ve, so velocity is +ve.

But because slope decreasing so

therefore acceleration is

16)

Slope of curve B and slope of curve A will be same before time TB only.

17)

All the graphs shows that speed is increasing as acceleration is positive in all three graphs for
given time interval. Also we can say that area under a-t graph i.e. change in velocity is positive
so velocity is increasing.

18)
19)

20)

21)

observing the graph, we can extract the following.

22)
 23)

24)

25)

CHEMISTRY

26) Bohr can explain hydrogen-like elements.

27)

28)

29)

Conceptual

30)

For

31)

Conceptual

32)

Lyman shortest wavelength

Balmer series longest wavelength

33)

34)
35) Bohr's atomic model is valid for single electron species.

36)

frequency of revolution of electron

37) Decreases to a greater negative value

38)

39) For Balmer series

n1 = 2 & n2 = 3, 4, 5, 6, .....
3rd line of Balmer series from 5 2

40) In lyman series

41)
Therefore, x + y = 17
42)

43) When intensity is double, number of electrons emitted per second also doubled, but
average energy of photoelectrons emitted remain same.

44) 2nd Bohr orbit of H

For He+

45)

46)

Total lines

47)
 48)

49) Energy of one photon

∴ % of energy converted to K.E

50)

∴ Angular momentum

MATHEMATICS

51) Given, series 1 + 3 + 7 + 13 + 21 + .....

Also, tn = 9901 ....(i)
Let
and
On subtracting

( neglecting because terms not negative)

52) Let

On subtracting, we get

or S = 6
53)

Let ab = 4k2 a+b = 5k

54) First common term = 11, d1 = 4, d2 = 5

LCM = (d1, d2) = 20
So series of common term is S = 11, 31, 51, 71

55) G.P. : a, ar, .........

a = 3 and r = or 3 (rejected)

G.P. is 9, 3, 1, ......
⇒ C is not correct

56) We have,

If not possible.
Hence,

57) If a, b, c are in H.P.,

are also in A.P.

are in A.P

are in H.P.

58) The sequence is a G.P. with common ratio .

Now from ,
59) (4x)2 + (2y)2 + (z)2 – 2yz – 4xz – 8xy = 0

⇒
∴ 4x = 2y , 2y = z , z = 4x
∴ 4x = 2y = z = k

∴ x, y, z are in G.P.

60)

= 40

61) Let a be the first term and r be the common ratio of the given G.P. then,

Clearly,

62)

63)
64)

Put 1, 3, 5, 7 verification
1+7=8
3 × 5 = 15

65)

66)

(n - 2) 180 = n/2 [2a + (n-1)d]

67)

Here, are the terms in G.P. with

68)
⇒ a2b3c4 < 42.63.84

69)
 (other terms get cancelled)

70)

71) Tn = 1 + 2 + 3 + ..... + n

Now

72)

Also

Now

=6

73) n AM's between 3 & 54

3, A1, A2, ........ An, 54
8th mean = 3 + 8d
(n – 2)th mean = 3 + (n – 2)d
54 = 3 + (n + 1)d ..... (1)
( 54 is (n + 2)th term)

⇒ 15 + 40d = 9 + 3(n – 2)d

⇒

⇒ from eq. (1)

⇒ 6n + 6 = 153n – 2346
⇒ n = 16

74) We can write

S consists of just one element.

75) We have Therefore,

or
Hence, the least value is 6.